For the process

H₂O(l) $\to$H₂O(g)

 at T=100$°$C and 1 atmosphere pressure, the correct choice is:

1. $∆{\mathrm{S}}_{\mathrm{system}}>0 \mathrm{and} ∆{\mathrm{S}}_{\mathrm{surrounding}}>0$

2. $∆{\mathrm{S}}_{\mathrm{system}}>0 \mathrm{and} ∆{\mathrm{S}}_{\mathrm{surrounding}}<0$

3. $∆{\mathrm{S}}_{\mathrm{system}}<0 \mathrm{and} ∆{\mathrm{S}}_{\mathrm{surrounding}}>0$

4. $∆{\mathrm{S}}_{\mathrm{system}}<0 \mathrm{and} ∆{\mathrm{S}}_{\mathrm{surrounding}}<0$

 During an adiabatic process:

1. pressure is maintained constant

2. gas is isothermally expanded

3. there is perfect heat insulation

4. the system changes heat with surroundings

Heat of combustion $∆\mathrm{H}°$ for C(s), H₂(g) and CH₄(g) are -94, -68 and -213 kcal/mol. Then, $∆\mathrm{H}°$ for 

C(s) + 2H₂(g) $\to$CH₄(g) is                                                                 

1. -17 kcal/mol                         

2. -111 kcal/mol

3. -170 kcal/mol                       

4. -85 kcal/mol

If 50 calorie are added to a system and system does work of 30 calorie on surroundings, the change in internal energy of system is:

1. 20 cal                             

2. 50 cal

3. 40 cal                             

4. 30 cal

For the process:

H₂O(l)[1 bar, 373 K] $\rightleftharpoons$H₂O(g)[1 bar, 373 K] the correct set of thermodynamic parameters are:

1. $∆\mathrm{G}=0; ∆\mathrm{S}=+\mathrm{ve}$

2. $∆\mathrm{G}=0; ∆\mathrm{S}=-\mathrm{ve}$

3. $∆\mathrm{G}=+\mathrm{ve}; ∆\mathrm{S}=0$

4. $∆\mathrm{G}=-\mathrm{ve}; ∆\mathrm{S}=+\mathrm{ve}$

The internal energy change when a system goes from state A to B is 40 kJ/mol. If the system goes from A to B by a reversible path and returns to state A by an irreversible path. What would be the change in internal energy?

1. 40 kJ                                   

2. >40 kJ

3. <40 kJ                                 

4. Zero

Change in entropy is negative for:

1. Bromine (l)$\to$Bromine(g)

2. C(s) + H₂O(g) $\to$CO(g) + H₂(g)

3. N₂(g,10 atm)$\to$N₂(g,1 atm) 

4. Fe ( 1mol, 400 K) $\to$ Fe( 1mol, 300 K)

The mathematical form of the first law of thermodynamics when heat (q) is supplied and W is work done by the system (-ve) is:

1. $∆$U=q+W                       

2. $∆$U=q-W

3. $∆$U=-q+W                       

4. $∆$U= -q-W

Which statements are correct?

1. $2.303 \log \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\frac{∆{\mathrm{H}}_{\mathrm{vap}.}}{\mathrm{R}}\frac{\left[{\mathrm{T}}_2-{\mathrm{T}}_1\right]}{{\mathrm{T}}_1{\mathrm{T}}_2}$ is called Clausius-Clapeyron equation

2. $\frac{∆{\mathrm{H}}_{\mathrm{vap}.}}{\mathrm{Boiling} \mathrm{Point}}=88 J mo{l}^{-1}{K}^{-1}$ is called Trouton's rule

3. Entropy is a measure of unavailable energy, i.e.,

     unavailable energy = entropy x temperature

4. All of the above

$∆\mathrm{S}°$ will be highest for the reaction:

1. Ca(s) + 1/2 O₂(g) $\to$CaO(s)

2. CaCO₃(s)$\to$CaO(s) + CO₂(g)

3. C(s) + O₂(g)$\to$CO₂(g)

4. N₂(g) + O₂(g)$\to$2NO(g)